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Find the causal inverse of $$h[n]=\delta[n]-\alpha\delta[n-1]$$

we have $h[0]=1$ and $h[1]=-\alpha$ also $h[n]=0$ for $n>1$

From the formula $$ h_i[n]=\sum_{i=1}^n\frac{h[n]h_i[n-i]}{h[0]} $$

we should have the recursive difference equation $$ h_i[n]=-\alpha h_i[n-1] $$ However this result is different from the book Digital Signal Processing- (Proakis) book solution

Also notice that the book stated that $h[n]=0$ for $n≥\alpha$ which does not make sense to me

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  • $\begingroup$ In general it is useful to add the title and edition of a book you quote. Proakis wrote quite a few books. $\endgroup$
    – Matt L.
    Nov 27, 2021 at 16:48

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Of course it should be $n\ge 2$, that's a typo in your edition. The sign of the formula in your question is wrong. It should be

$$h[0]h_I[n]=-\sum_{k=1}^nh[k]h_I[n-k],\qquad n>0\tag{1}$$

With $h[0]=1$, $h[1]=-\alpha$, and $h[n]=0$ for $n>1$, Eq. $(1)$ simplifies to

$$h_I[n]=-h[1]h_I[n-1]=\alpha h_I[n-k],\qquad n>0\tag{2}$$

And since $h_I[0]=1/h[0]=1$ you obtain the result given in the book.

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  • $\begingroup$ Oh, it means that the formula on the book has also a typo. Since for $n≥1$, we have $h_I[n]*h[n]=0$ so by taking a single term of that convolution sum to the right, we get the missing negative sign? $\endgroup$
    – 12775
    Nov 28, 2021 at 1:53
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    $\begingroup$ @qcpz: that's right, by splitting up the sum and equating it to zero for $n>0$ you get the negative sign on one side of the equation. $\endgroup$
    – Matt L.
    Nov 28, 2021 at 12:58

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